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What is the binomial expansion of (x + 2y)??

O 2x + 14x5y + 42x³y² + 70x¹y³ +70x³y² + 42x²y³ + 14xy + 2y²
O x² + 14xy + 42x³y² + 70x¹y³ +70x³y² + 42x²y³ + 14xy + y²
O x² + 7xy + 21x³y² + 35x4y³ + 35x³y² + 21x²y5 + 7xy + y²
O x² + 14x5y + 84x³y2 + 280x¹y³ +560x³y² +672x²y³ + 448xy + 128y²

User Josh M
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Answer: The binomial expansion of (x + 2y) can be found using the binomial theorem, which states that for any positive integer n: (x + y)^n = C(n,0) * x^n * y^0 + C(n,1) * x^(n-1) * y^1 + C(n,2) * x^(n-2) * y^2 + ... + C(n,n) * x^0 * y^n Here, C(n,k) represents the binomial coefficient, which can be calculated using the formula: C(n,k) = n! / (k! * (n-k)!) Let's apply this formula to the expression (x + 2y): For n = 1: (x + 2y)^1 = C(1,0) * x^1 * (2y)^0 + C(1,1) * x^0 * (2y)^1 = 1 * x + 1 * 2y = x + 2y For n = 2: (x + 2y)^2 = C(2,0) * x^2 * (2y)^0 + C(2,1) * x^1 * (2y)^1 + C(2,2) * x^0 * (2y)^2 = 1 * x^2 + 2 * x * 2y + 1 * (2y)^2 = x^2 + 4xy + 4y^2 As you can see, the binomial expansion of (x + 2y) is: (x + 2y)^n = x^n + (nC1) * x^(n-1) * 2y + (nC2) * x^(n-2) * (2y)^2 + ... + (nCn) * 2y^n It's important to note that the binomial theorem is applicable for any positive integer n. The coefficients and powers of x and y change depending on the value of n.

User Teresa
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